One-point Green's function Why is one-point Green's function for scalar field equal to zero? Can one prove it using path integral formalism? 
 A: It is not necessarily zero. The one-point Green functions correspond to the mean value of the field $\langle \phi(x)\rangle$, which one could naively assume to vanish if the theory is invariant under $\phi\to-\phi$ (which need not to be the case). But even if the action is invariant under this transformation, $\langle \phi(x)\rangle$ can be non-zero if the symmetry is spontaneously broken.
A: As Adam said this is not always true. This statement is only correct if you are in the ground state. If you want a path integral formalism demonstration, you can write a partition function introducing a counting field J in your action:
$$
Z[J]=\int \mathcal D\phi \exp\left(-i\int dt \phi(t) A \phi(t)+2i\int\phi(t)J(t)\right)
$$
with A the propagator. if we introduce C the Green function so that $AC(t,t')=\delta(t-t')$, we can  make the chage of variable:
$$\tilde\phi(t)=\phi(t)-i\int dt'C(t,t')J(t')$$
you get 
$$
Z[J]=\int \mathcal D\phi \exp\left(-i\int dt \tilde\phi(t) A \tilde\phi(t)+
-i\iint dt dt' J(t)C(t,t')J(t')\right)
$$
so that:
$$
\langle\phi(\tau)\rangle=\left.\frac{\delta \ln Z[J]}{i\delta J}\right|_{J=0}=-2i\lim_{J\to0}\int dt'C(\tau,t') J(t')=0
$$
